NEET 2025 · ChemistryPrevious Year Question
Consider the following compounds: \ce{KO2}, \ce{H2O2} and \ce{H2SO4}. The oxidation states of the underlined elements (\ce{K} in \ce{KO2}, \ce{O} in \ce{H2O2}, \ce{S} in \ce{H2SO4}) in them are, respectively,
- A.(a) $+1$, $-2$ and $+4$
- B.(b) $+4$, $-4$ and $+6$
- C.(c) $+1$, $-1$ and $+6$✓
- D.(d) $+2$, $-2$ and $+6$
Correct Answer
(C) (c) $+1$, $-1$ and $+6$
Solution & Explanation
\textbf{Answer:} (c) $+1$, $-1$ and $+6$. \textbf{Solution:} \ce{KO2} is potassium superoxide, in which \ce{K} is $+1$ (each \ce{O} is $-1/2$). \ce{H2O2} is a peroxide, so \ce{O} is $-1$. In \ce{H2SO4}: $2(+1) + S + 4(-2) = 0$, giving \ce{S} $= +6$. Hence the underlined elements are $+1$ (\ce{K}), $-1$ (\ce{O} in peroxide) and $+6$ (\ce{S}), i.e. option (c).
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